1. a) If a particle's position is given by LaTeX: x\:=\:4-12t\:+\:3t^2x = 4 − 12 t + 3 t 2(where t is in seconds and x is in meters), what is its velocity at LaTeX: t=1st = 1 s? b) Is it moving in the positive or negative direction of LaTeX: xx just then? c) What is its speed just then? d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculations.) e) Is there ever an instant when the velocity is zero? If so, give the time LaTeX: tt ; if not, answer no. f) Is there a time after LaTeX: t=3st = 3 s when the particle is moving in the negative direction of LaTeX: xx? If so, give the time LaTeX: tt; if not, answer no. (Hint: Speed= LaTeX: \mid v\mid∣ v ∣)

Answer:

Explanation:

We shall apply Newton's laws of cooling .

dT / dt = k ( T₁ - T₂ )

dT is change in temperature in time dt , k is constant , T₁ and T₂ are temperature of cold  object and hot surrounding respectively.

For  the first case of 4 minutes

(14 - 10)/ 4 = k [(14+10) /2 - 38 ]  , T₁ = (14+10) /2 , the average  temperature

1 = - 26 k

Using the formula for next two minutes

T-14 / 2 = k [(14+T) /2 - 38 ]

T-14 / 2 = 1/26[ 38 - (14+T) /2 ]

T / 2 - 7 = 1.46 - .27 - T / 52

T/2 + T/52 = 7 + 1.46 - .27

= .52 T = 8.19

T = 15.75 degree

Answer:

The kinetic energy of the merry-go-round is [tex]\bf{475.47~J}[/tex].

Explanation:

Given:

Weight of the merry-go-round, [tex]W_{g} = 826~N[/tex]

Radius of the merry-go-round, [tex]r = 1.17~m[/tex]

the force on the merry-go-round, [tex]F = 57.8~N[/tex]

Acceleration due to gravity, [tex]g= 9.8~m.s^{-2}[/tex]

Time given, [tex]t=3.47~s[/tex]

Mass of the merry-go-round is given by

[tex]m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg[/tex]

Moment of inertial of the merry-go-round is given by

[tex]I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}[/tex]

Torque on the merry-go-round is given by

[tex]\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m[/tex]

The angular acceleration is given by

[tex]\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}[/tex]

The angular velocity is given by

[tex]\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}[/tex]

The kinetic energy of the merry-go-round is given by

[tex]E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J[/tex]

The given diagram of the magnet shows attraction which occurs between opposite poles therefore, A and C could both be the north poles. Thus, the correct option is C.

A magnet is a material or an object which produces a magnetic field around itself. The magnetic field of this magnet is invisible however it is responsible for the most notable property of a magnet which is a force that is responsible for pulling on other ferromagnetic materials, such as iron, steel, nickel, cobalt, etc. towards it and attracts or repels other magnets as well.

Some important properties of magnet include a magnet has two poles which are north pole and south pole, magnets attract ferromagnetic materials, the similar poles of two magnets repel each other, whereas the opposite poles of two magnets attract each other. A magnet hung up in the air always comes to rest facing the north-south direction. The poles of a magnet are arranged in pairs.

Therefore, the correct option is C.

Learn more about Magnet here:

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Given that,

Frequency of radio wave is from  f = 10⁵ Hz and f' = 10³ Hz

We need to find the wavelength of radio wave. The speed of a wave is given by formula as follows :

If f = 10⁵ Hz

[tex]v=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{10^5}\\\\\lambda=3000\ m[/tex]

or

[tex]\lambda=3\ km[/tex]

If f' = 10³ Hz

[tex]v=f\lambda\\\\\lambda'=\dfrac{c}{f'}\\\\\lambda'=\dfrac{3\times 10^8}{10^3}\\\\\lambda'=300000\ m[/tex]

or

[tex]\lambda'=300\ km[/tex]

So, the wavelength of this radio wave on the order of kilometres.

The wavelength of this radio wave is on the order of meters.

The wavelength of a radio wave can be calculated using the formula: wavelength = speed of light / frequency. In this case, the given frequency of the radio wave is on the order of 10^5 Hz. To compare it with the lowest audible sound waves with a frequency of 20 Hz, we can use the formula again. Taking the speed of sound in air at 20 °C as 343 m/s, we can calculate the wavelength of the lowest audible sound waves. By comparing the wavelengths of the two waves, we can determine the order of magnitude of the wavelength of the radio wave.

Let's calculate the wavelengths:

Wavelength of radio wave = speed of light / frequency = (3 x 10^8 m/s) / (10^5 Hz) = 3000 m

Wavelength of lowest audible sound wave = speed of sound / frequency = (343 m/s) / (20 Hz) = 17.15 m

Comparing the two wavelengths, we can see that the wavelength of the radio wave is on the order of meters.

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According to the law of conversion of mass, the mass of a closed system is neither made or taken away in a physical change. It is the same for a chemical change as well.

Therefore, the mass of the substance remains 8g.

Best of Luck!

Answer:

y = 0.1sin(4.5x - 345.6t)

Explanation:

Parameters given:

Length of 5 wavelengths = 7 m

Length of one wavelength, λ = 7/5 = 1.4 m

Frequency of wave, f = 55 Hz

Peak to Valley displacement = 20 cm = 0.2 m

Amplitude is half of the Peak to Valley displacement = 0.1 m

The wave function of a wave traveling in the positive x direction is given generally as:

y = Asin(kx - wt)

Where A = amplitude

k = Wave factor = 2π/λ

x = displacement on the x axis

w = angular frequency = 2πf

t = time taken

=> Wave factor, k = 2π/λ = 2π/(1.4) = 4.5 m^(-1)

=> Angular frequency, w = 2πf = 2π * 55 = 345.6 Hz

Therefore, inserting all the necessary parameters, we get that the wave function is:

y = 0.1sin(4.5x - 345.6t)

Answer:

The solution to the question above is explained below:

Explanation:

For which solid is the lumped system analysis more likely to be applicable?

Answer

The lumped system analysis is more likely to be applicable for the body cooled naturally.

Question :Why?

Answer

Biot number is proportional to the convection heat transfer coefficient, and it is proportional to the air velocity. When Biot no is less than 0.1 in  the case of natural convection, then lumped analysis can be applied.

Further explanations:

Heat is a form of energy.

Heat transfer describes the flow of heat across the boundary of a system due to temperature differences and the subsequent temperature distribution and changes. There are three different ways the heat can transfer: conduction, convection, or radiation.

Heat transfer  analysis which utilizes this idealization is known as the lumped system analysis.

The Biot number is a criterion dimensionless quantity used in heat transfer calculations which gives a direct indication of the relative importance of conduction and convection in determining the temperature history of a body being heated or cooled by convection at its surface. In heat transfer analysis, some bodies are observed to behave like a "lump" whose entire  body temperature remains essentially uniform at all times during a heat transfer process.

Conduction is the transfer of energy in the form of heat or electricity from one atom to another within an object and conduction of heat occurs when molecules increase in temperature.

Convection is a transfer of heat by the movement of a fluid. Convection occurs within liquids and gases between areas of different temperature.

Final answer:

The lumped system analysis is more applicable to the solid cooled naturally because natural convection is slower, allowing the solid's temperature to equalize, which suits the assumptions of lumped system analysis better than the enhanced convection caused by a fan.

Explanation:

The lumped system analysis is more likely to be applicable to the solid that is being cooled naturally rather than the one with a fan. The reason for this is that the use of a fan increases the flow of air over the surface, enhancing convection and therefore increasing the rate of heat transfer. This forced convection reduces the likelihood that the temperature gradient within the solid will remain uniform, which is a key assumption of lumped system analysis – that the object's temperature is spatially uniform and can be modeled as a lumped capacitance.

In contrast, when cooling occurs naturally, the heat transfer is driven by natural convection, which tends to be slower as it relies on the thermal expansion of the fluid around the solid due to the heat transfer from the solid. The slower rate of heat transfer allows the temperature within the solid to equalize more readily, making lumped system analysis a more appropriate simplification for predicting the thermal response of the solid.

The spacecraft had an angular acceleration of 0.00873 rad/s². At 6 minutes into the flight, a point on the skin of the spacecraft had a radial (centripetal) acceleration of 0.01167 m/s² and a tangential acceleration of 0.0371 m/s².

First, let's find the angular acceleration which is the rate of change of angular velocity. The spacecraft went from 0 to 1 revolution per minute in 12 minutes. This is an angular acceleration (α) of 1 rev/min/12 min = 1/12 rev/min². However, we generally measure angular acceleration in rad/s², not rev/min². We know that 1 revolution = 2π rad and 1 min = 60 s, so we can convert our units to get α = ((1/12 rev/min²) x (2π rad/rev) x (1 min/60 s)²) = 0.00873 rad/s².

Next, we find the radial (centripetal) and tangential components of linear acceleration at 6 minutes. At 6 minutes, the angular velocity (ω) is (α x t) = ((1/12 rev/min) x 6 min) = 0.5 rev/min. Convert this to rad/s: ω = ((0.5 rev/min) x (2π rad/rev) x (1 min/60 s)) = 0.05236 rad/s. Now, the radial or centripetal acceleration (a_r) is given by (ω² x r) and the tangential acceleration (a_t) is given by (α x r), where r is the radius of the spacecraft, which is half of the diameter or 8.5 m / 2 = 4.25 m. We can plug the numbers in to get: a_r = (ω² x r) = (0.05236 rad/s)² x 4.25 m = 0.01167 m/s² and a_t = α x r = 0.00873 rad/s² x 4.25 m = 0.0371 m/s². So, the radial acceleration is 0.01167 m/s² and the tangential acceleration is 0.0371 m/s².

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Answer:

12.27 ft/s

Explanation:

At 72 ft above the ground,  the balloons height increases at a rate of 4ft/s. For 66s, vertical distance moved, y = 4ft/s × 66 s = 264 ft. When the balloon is at 72 ft above the ground, just below it, the bicycle which moves at a rate of 12 ft/s in 66 s, covers a horizontal distance, x = 12ft/s 66 = 792 ft.

The distance between the bicycle and the balloon 66 s later is given by

s = √(x² + (y + 72)²) = √(792² + (264 + 72)²) = √(792² + 336²) = √740160 ft = 860.33 ft

From calculus

The rate of change of the distance between the balloon and bicycle s is obtained by differentiating s with respect to t. So,

ds/dt = (1/s)(xdx/dt + ydy/dt)

dx/dt = 12 ft/s, x = 792 ft, dy/dt = 4 ft/s, y = 264 ft, s = 860.33. These are the values of the variables at t = 66 s.

So, substituting these values into ds/dt, we have

ds/dt = (1/860.33)(792 ft × 12 ft/s + 264 ft × 4ft/s) = (1/860.33)(9504 + 1056) = 10560/860.33 = 12.27 ft/s

Answer:

Explanation:

check the pictures attached to further understand and i hope it works

Answer:

[tex]v = \frac{-384}{5} (1 - \frac{12}{5} e^{ -5t/12})[/tex]

Explanation:

Weight of the firefighter, W = 192 lb

W = mg

g = 32 ft/s²

Mass of the firefighter, m = W/g

m = 192/32

m = 6 slugs

k = 2.5 lb-s/ft

The force, F = ma = kv

2.5v = 6a

a = 2.5v/6

a = 5v/12

The fundamental dynamic equation;

dv/dt + drag + gravity = 0

dv/dt = -g-a

dv/dt = -32-5v/12..............(a)

The motion will attain terminal velocity when dv/dt = 0

-32 - 5v/12 = 0

-32 = 5v/12

-384 = 5v

v₀ = -384/5

v₀ = -384/5

dv/dt = -32 - 5v/12

[tex]\frac{dv}{-32 -5v/12} = dt[/tex]

[tex]-12/5 ln(32 + 5v/12) = t + c\\ln(32 + 5v/12) = -5t/12 + lnc\\ln(32 + 5v/12) - ln c = -5t/12\\ln\frac{32 + 5v/12}{c} = -5t/12[/tex]

Take exponential of both sides

[tex]\frac{32 + 5v/12}{c} =e^{ -5t/12}\\32 + 5v/12 = ce^{ -5t/12}\\5v/12 = -32 + ce^{ -5t/12}\\v = 12/5 (-32 + ce^{ -5t/12})\\[/tex]

c = v₀ = -384/5

[tex]v = 12/5 (-32 - \frac{384}{5} e^{ -5t/12})\\v = \frac{-384}{5} (1 - \frac{12}{5} e^{ -5t/12})[/tex]

Answer:

Only technician A is correct

Explanation:

Analysis of technician A's statements

The refrigeration cycle contains four major components: the compressor, condenser, expansion device, and evaporator. Refrigerant remains piped between these four components and is contained in the refrigerant loop.

The refrigerant begins as a cool vapor and heads to the first component:

1. the compressor

It forces the refrigerant through the system. In the process of being compressed the cool, gaseous refrigerant is turned to a very hot and high-pressure vapor.

Analysis of technician B's statement

The condenser’s job is to cool the refrigerant so that it turns from a gas into a liquid, or condenses. This happens when warm outdoor air is blown across the condenser coil that is filled with hot, gaseous refrigerant.

The evaporator is responsible for cooling the air going to the space by boiling (evaporating) the refrigerant flowing through it. This happens when warm air is blown across the evaporator as cold refrigerant moves through the evaporator coil. Heat transfers from the air to the refrigerant, which cools the air directly before it is vented to the space.

Answer:

7.54 x[tex]10^{-7}[/tex] C

Explanation:

The complete question is:

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each 19 cm in diameter, separated by 1.0 cm.How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3×106N/C of the field causes a spark.

SOLUTION:

As you know that, parallel plate capacitance can be defined as

C=εo A/d

Where,

εo is the permittivity of free space constant, A is area of the capacitor plates, and d is the distance between them

Also, C= Q/V

where,

'Q' defines charge stored on the capacitor and 'V' express the potential difference between the plates.

Equating the equations.

εo A/d =  Q/V

Q= (εo A V)/d

for a uniform electric field (such as the one between the plates of a parallel-plate capacitor)

V= Ed ->where 'E' defines magnitude of the electric field.

Therefore,

Q= (εo A E d)/d  =>εo A E ->eq(1)

The area of the plates is given by

A= πr² => π(d/2)² =>π(0.19/2)

A=0.0284m²

Given: E= 3 x [tex]10^{6}[/tex] N/C

Substituting all the required values in eq(1)

(1)=> Q= (8.85x[tex]10^{-12}[/tex]) (-.0284)(3 x [tex]10^{6}[/tex] )

Q= 7.54 x[tex]10^{-7}[/tex] C

Therefore,7.54 x[tex]10^{-7}[/tex] C can be added to each of the plates before a spark jumps between the two plates

Answer:

kinetic energy does not change

Explanation:

you can use the formula for the kinetic energy of the electron and for the radius of the trajectory of an electron in a uniform magnetic field:

[tex]E_k=\frac{1}{2}m_ev^2\\\\r=\frac{m_e v}{qB}[/tex]

me: mass of the electron

B: magnetic field

q: charge of the electron

r: radius

By doing v the subject of the formula and replace it in the expression for the kinetic energy you obtain:

[tex]v=\frac{rqB}{m_e}\\\\E_k=\frac{1}{2}m_e(\frac{rqB}{m_e})^2=\frac{r^2q^2B^2}{2m_e}[/tex]

as youcan see, all parameters r, q, B and me are constant.

hence, the kinetic energy does not change

Answer:

 k = 2.65 10² N / m

Explanation:

As the pellets shoot up we can use energy conservation

Starting point. Compressed spring rifle

            Em₀ = Ke = ½ k x²

Final point. Highest point of the path

            [tex]Em_{f}[/tex] = U = m g y

Energy is conserved

            Em₀ = Em_{f}

           ½ k x² = m g y

           k = 2 m g y / x²

Let's calculate

            k = 2  1.06 10⁻²  9.8  6.26 / (6.61 10⁻²)²

             k = 2.65 10² N / m

Answer:

Answer in explanation

Explanation:

In this question, we will be providing an explanation as to why OPMD provides more than 10^10 higher rate acceleration than CA

we proceed as follows;

Kcat is a measure of the catalytic activity of an enzyme indicating how many reactions a molecule of enzyme can catalyze per second. But Michaelis constant, Km also plays role during rate acceleration. The kcat/Km ratio is useful for comparing the activities of different enzymes. It is also possible to assess the efficiency of an enzyme by measuring its catalytic proficiency.

This value is equal to the rate constants for a reaction in the presence of the enzyme (kcat/Km) divided by the rate constant for the same reaction in the absence of the enzyme (kn). an enzyme with rapid binding might evolve a mechanism that favored a faster reaction.

An electric field around the OMPD active site enhances the rate of formation of the ES complex. Electrostatic effects allow OMPD to bind and remove OMPD much faster than expected from random collisions of enzyme and substrate.

Answer:

  L = 2.98 m

Explanation:

In this exercise we are told that there is no friction, so we can use energy conservation

Starting point. Compressed spring

        Em₀ = Ke = ½ k x²

Final point. At the highest point of the ramp

        [tex]Em_{f}[/tex] = U = mg h

As there is no friction the energy is conserved

         Emo = Em_{f}

         ½ k x² = mg h

         h = ½ k x² / mg

         h = ½ 400 0.50² / (5.0 9.8)

         h = 1.02 m

This is the height that the body reaches, to calculate the distance traveled on the ramp let's use trigonometry

          sin θ = h / L

         L = h / sin θ

         L = 1.02 / sin 20

         L = 2.98 m

Tt will take approximately 0.14 seconds for the force of 2.5 N to bring the 0.25 kg object to rest from an initial velocity of 1.4 m/s

The question involves the application of Newton's second law of motion and the concepts of force, mass, acceleration, and time to bring an object to rest. Given a 0.25-kg object moving at 1.4 m/s, and a force of 2.5 N acting on it, we can first determine the acceleration using Newton's second law, F=ma, which gives us an acceleration of 10 m/s2. Next, we use the kinematic equation v = u + at (where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time) to find the time. Since the object comes to rest, v = 0 m/s, solving the equation for t gives us a time of 0.14 seconds to bring the object to rest.

Answer:

triple covalent bond

Explanation:

A triple covalent bond is formed when three pairs of electrons (six electrons) are shared between the two combining atoms. A triple bond is shown by marking three short lines between the two symbols of the atoms. It requires three more electrons to attain the stable octet.

- Hope this helps! If you need a further explanation please let me know.

Final answer:

A triple covalent bond connects the two nitrogen atoms in this molecule N2.

Explanation:

A triple covalent bond connects the two nitrogen atoms in this molecule. Nitrogen atoms have 5 valence electrons and need to share 3 pairs of electrons to achieve a valence octet. The triple bond between the nitrogen atoms allows them to share 3 pairs of electrons and form a stable diatomic molecule, N2.

Answer:

The answer to your question is Work = 784.8 Joules

Explanation:

Data

mass = 20 kg

height = 4 m

time = 10 s

Mechanic work is defined as the force applied to a body so it moves a distance.

Formula

    Work = force x distance

Process

1.- Calculate the force

Force = mass x gravity

-Substitution

Force = 20 x 9.81

-Result

Force = 196.2 N

2.- Calculate the Work

Work = 196.2 x 4

-Result

Work = 784.8 Joules

Answer:

Measurement and Direction

Explanation:

Displacement is a vector quantity because it has magnitude (the measure o displacement)

Answer:

Measurement and direction.

Explanation:

Answer:

2.36 x 10^5 kg

Explanation:

radius of hose, r = 0.017 m

radius of underground pipe, R = 0.088 m

number of hoses, n = 3

velocity of water in underground pipe, V = 2.7 m/s

Let v is the velocity of water in each hose.

According to the equation of continuity

A x V = n x a x v

π R² x V = n x π x r² x v

0.088 x 0.088 x 2.7 = 3 x 0.017 x 0.017 x v

v = 24.12 m/s

(a) Amount of water poured onto a fire in one hour by all the three hoses              

                    = n x a x v x density of water x time

                    = 3 x 3.14 x 0.017 x 0.017 x 24.12 x 1000 x 3600

                    = 2.36 x 10^5 kg

Thus, the amount of water poured onto the fire in one hour is 2.36 x 10^5 kg.

a) Using the principles of fluid dynamics, specifically the continuity equation, we found that the velocity of water in the hoses is 2.02 m/s. The volume flow rate in the hoses is 0.0055 m³/s. Therefore, the mass of water poured onto a fire in one hour by all three hoses is 19800000 kg.

a) To answer this question, we need to apply the principles of fluid dynamics, specifically the concept of continuity equation which states that the mass flow rate (the mass of water passing through a given spot per unit time) through a pipe system is constant.

First, let's calculate the cross sectional area of the hydrant pipe (A1) which is πr² = π(0.088m)² ≈ 0.0243 m².

Then, calculate the area of one fire hose (A2) which is π(0.017m)² ≈ 0.000907 m².

Since there are three hoses, the total area for the three hoses (A3) is 3 * A2 ≈ 0.00272 m².

Then, we apply the continuity equation, which is

A1v1=A3v3. We can solve for v3 (the velocity of the water in the hoses), v3 = A1v1/A3 ≈ 2.02 m/s.

To find the mass of water delivered in one hour, we first find the volume flow rate in the hoses,

Q= A3v3 = 0.00272 m² * 2.02 m/s ≈ 0.0055 m³/s.

The volume of water in one hour is Q * 3600s = 19800 m³.

Then, since water has a density (ρ) of 1000 kg/m³, the mass (m) of the water is ρ*volume which equals

19800 * 1000 kg = 19800000 kg.

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Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

[tex]Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3][/tex]

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

[tex]F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3][/tex]

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

Answer: C = Q/4πR

Explanation:

Volume(V) of a sphere = 4πr^3

Charge within a small volume 'dV' is given by:

dq = ρ(r)dV

ρ(r) = C/r^2

Volume(V) of a sphere = 4/3(πr^3)

dV/dr = (4/3)×3πr^2

dV = 4πr^2dr

Therefore,

dq = ρ(r)dV ; dq =ρ(r)4πr^2dr

dq = C/r^2[4πr^2dr]

dq = 4Cπdr

FOR TOTAL CHANGE 'Q', we integrate dq

∫dq = ∫4Cπdr at r = R and r = 0

∫4Cπdr = 4Cπr

Q = 4Cπ(R - 0)

Q = 4CπR - 0

Q = 4CπR

C = Q/4πR

The value of C in terms of Q and R is [Q/4πR]

1) 2 Forces

2) 563 N, upward

3) [tex]8.66 m/s^2[/tex], upward

4) Speed is decreasing

Explanation:

1)

There are two forces acting on teacher during his skydiving:

- The force of gravity (also known as weight), acting downward (towards the Earth's centre), of magnitude

[tex]W=mg[/tex]

where

m is the mass of the teacher

g is the acceleration due to gravity

- The air resistance, of magnitude R = 1200 N, acting upward (air resistance always acts in the direction opposite to the direction of motion)

There are no other forces, therefore the correct answer is:

2 Forces

2)

The net force on the system is given by the vector resultant of the two forces described in part 1). Since the two forces act along the same line but in opposite directions, the net force will be equal to the difference between the two forces, so:

[tex]F_{net}=W-R=mg-R[/tex]

where we have chosen downward as positive direction, and where

m = 65 kg is the mass of the teacher

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

R = 1200 N is the air drag

Therefore, the net force on the teacher is

[tex]F_{net}=(65)(9.8)-1200=-563 N[/tex]

And the negative sign means that the direction of the net force is upward.

3)

The acceleration of the system can be found by using Newton's second law of motion, which can be written as

[tex]a=\frac{F_{net}}{m}[/tex]

where

a is the acceleration

[tex]F_{net}[/tex] is the net force

m is the mass of the teacher

Here we have

[tex]F_{net}=-563 N[/tex]

m = 65 kg

Therefore, the acceleration is

[tex]a=\frac{-563}{65}=-8.66 m/s^2[/tex]

And the negative sign indicates that the direction of the acceleration is opposite to the motion (so, upward).

4)

From part 3), we observed that:

- The motion of the teacher is downward (he is moving downward)

- However, the direction of his acceleration is upward

This means that the velocity and the acceleration of the system have opposite directions.

As we know, the velocity of an accelerating system can be written as

[tex]v=u+at[/tex]

where

v is the velocity after time t

u is the initial velocity

a is the acceleration

Since a is negative, we see that [tex]at<0[/tex], so [tex]v<u[/tex]: this means that the velocity of the teacher is decreasing with time, so, the teacher is slowing down, and therefore his speed is decreasing.

(a) Two forces are acting on the system.

(b) The magnitude of net force on the system is 592.4 N.

(c) The magnitude of the acceleration is 9.11 m/s².

(d) The speed of the system decreases with time

The given parameters;

The forces acting on the teacher are downward force due to the teachers weight and drag force acting upward opposing the downward motion.

The magnitude of net force on the system is calculated as follows;

[tex]F_{net}= W - F_f\\\\F_{net}= (65 \times 9.8) - (1200)\\\\F_{net} = -592.4 \ N[/tex]

The magnitude of the acceleration is calculated as follows;

[tex]a = \frac{-592.4}{65} \\\\a = -9.11 \ m/s^2\\\\|a| = 9.11 \ m/s^2[/tex]

The speed of the system is calculated as follows;

[tex]v= u + at\\\\v= 0 + (-9.11)t\\\\v = -9.11t[/tex]

The speed of the system decreases with time.

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Answer:

C) 4.4

Explanation:

The potential of a cylindrical capacitor is given by the formula:

[tex]V=\frac{2kq}{L\epsilon}ln(\frac{a}{b})\\\\\epsilon=\frac{2kq}{LV}ln(\frac{a}{b})[/tex]

where:

k : Coulomb Constant

L : length of the capacitor

a : outer radius

b : inner radius

V : potential

By replacing we obtain:

[tex]\epsilon=\frac{2(8.89*10^{9}N/m^2C^2)(8.8*10^{-6}C)}{(1m)(6000V)}ln(\frac{1.3mm}{1.1mm})=4.35[/tex]

Hence, the answer is C) 4.4 (4.35 is approximately 4.4)

hope this helps!!

Answer:

B

Explanation:

Answer:

Work done

Explanation:

The work done on an object is given as the product of force and distance traveled by the object.

J = Ft (Ns)

The change in the momentum of an object is equal to the impulse experienced by the object.

The work done on an object is given as the product of force and distance traveled by the object.

W = Fd (J)

The work done on an object causes change in the position of the object.

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Answer:

Will not.

Explanation:

just got the question correct.

Answer:

will not .

Explanation:

Answer:

saturated solution is the correct answer.

Explanation: